So there is an edge-colouring of a complete graph on R (the reals), with countably many colours that as no monochromatic triangle. To find it map R to (0,1) write the numbers in binary and if 2 numbers differ 1st in the kth digit use colour k.

Now this colouring has cycles of length 4. (1/4, 3/4, 1/3, 2/3 for example). You can get rid of cycles of length 4 by considering the 1st 2 binarary digits in which 2 numbers differ (and of course seperate colours if they only differ in 1 digit). Anyway my question is can we avoid cycles completely? i.e. does there exist a colouring of the complete graph on R such that there is no monochromatic cycle.

If the color of an edge comes from the pair of first binary differences, wouldn't (1/5, 4/5, 1/6, 5/6) give a monochromatic 4-cycle? Or did I misunderstand your construction?
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Tracy HallAug 17 '10 at 9:20

@Tracy: The 1st construction I gave has no 3 cycles but as you say does have 4-cycles. The question was weather there is a "better" colouring which does not have any cycles.
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Jonathan KarivAug 17 '10 at 15:56

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@Jonathan: I was talking about the second, supposedly 4-cycle-free construction. The 4-cycle (1/5, 4/5, 1/6, 5/6) alternates between binary expansions starting .00 and .11, and so as I understood your edge-coloring each step would be colored {1,2}: a monochromatic 4-cycle. This seems to be a fundamental problem with the approach: if there is some pair of numbers $p$ and $q$ such that the rule coloring the edge from $p$ to $q$ depends on only the first digits (or neglects any digit at all), then changing a neglected digit of $p$ and a neglected digit of $q$ yields a monochromatic 4-cycle.
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Tracy HallAug 17 '10 at 17:11

Thank you very much for this reference ! I am trying to read the beginning of this paper [1], and I am stuck somewhere though. They are splitting all the intervals according to some notion of "length" (the $G_n$), then taking the union of all of them. Well, I do not get why there are not as many $G_n$ as $\omega_1$ :-/ Or does it mean that an initial section of a well ordered set of power $\omega_1$ is of cardinal at most $\omega_0 ? If you think some reading may be fitting in this case.. Thanks ! :-) [1] ams.org/bull/1943-49-06/S0002-9904-1943-07954-2/…
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Nathann CohenAug 17 '10 at 10:29

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Nathann, every initial segment of the ordinal $\omega_1$ is countable, if this is what you are asking. This is because $\omega_1$ is the smallest uncountable ordinal, by definition.
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Joel David HamkinsAug 17 '10 at 11:28

Let me add to the above answer by noting that a stronger result holds and that, in the absence of the Continuum Hypothesis, the problems with the attempted construction of a 4-cycle-free coloring are unavoidable.

To fix notation, if $X$ is a set, $[X]^2$ is the set of 2-element subsets of $X$. We show that any coloring $c:[\omega_2]^2 \rightarrow \omega$ has a monochromatic 4-cycle. This will also give a shorter proof of the negative direction of the Erdos-Kakutani result mentioned above.

Erdos and Kakutani ask at the end of that beautiful 1943 paper whether if CH fails, and the sets $M_n$ consist of rationally independent numbers, then $\bigcup M_n$ has inner measure 0. Was this ever answered?
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AvshalomMay 10 at 10:22

@Avshalom This looks true to me. If you ask this as a separate question I will try to check the details and write an answer.
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AshutoshMay 10 at 21:28